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As an example I will use the Fresnel equations: $$t_{p,s}=\frac{2\cos\theta^i}{\cos\theta^{t,i}+\frac{n_2}{n_1}\cos\theta^{i,t}}\tag{T}$$

$$r_{p,s}=\frac{\cos\theta^{t,i}-\frac{n_2}{n_1}\cos\theta^{i,t}}{\cos\theta^{t,i}+\frac{n_2}{n_1}\cos\theta^{i,t}}\tag{R}$$

The subscripts on the LHS of $(\mathrm{T})$ & $(\mathrm{R})$; $p,s$ are linked to the superscripts $t,i$ on the RHS of $(\mathrm{T})$ & $(\mathrm{R})$. Where $p$ denotes the electric field component parallel to the plane of incidence (which is the page/screen), $s$ is the electric field perpendicular component to the plane of incidence. The superscripts $i$,$t$ denote incidence and transmission respectively.

At normal incidence $\theta^i=\theta^t=0$ and so $$r_p=r_s=r=\frac{E^r}{E^i}=\frac{n_1-n_2}{n_1+n_2}\tag{N}$$

If we consider the simple case of light incident normally from air ($n_1 \approx 1$) to glass ($n_2\approx 1.5$) then:

$$\color{blue}{\fbox{$\frac{E^r}{E^i}=-0.2$}}$$

Now this is where my main question begins: According to every source I encounter they typically state "the reason for the minus sign is because there is a phase difference of $\pi$ between the reflected wave and the incident wave."

Walter Lewin used the same calculation as in $(\mathrm{N})$ and reached the box I marked blue and asked his audience "What is the meaning of the minus?" Which can be seen at $53$ min in his Lecture "8.03 - Lect 18 - Index of Refraction, Reflection, Fresnel Equations, Brewster Angle". One of his students answered that the meaning of the minus is $180^{\circ}$ phase difference. Then Walter goes on to give further reinforcement that the minus means $\pi$ phase difference and gave an intuitive (heuristic) reasoning for this (which sadly does not answer my question here).


Here is a short extract from my lecture notes:

Lecture notes

The text I enclosed in red for those two bullet points requires justification. Why does 'positive' imply 'in-phase' and 'negative' imply 'phase difference of $\pi$'?


Finally, I would like to understand why the sign change occurs in the curve of $r_p$; as it is the only reflection to incident ratio that actually crosses the $\theta_i$ axis:

Lecture notes second page

Looking at the graph on the left I am told that the phase difference

$$\theta^r-\theta^i=\begin{cases} \pi & \text{if}\quad \theta^i \lt \theta_B \\ 0 & \text{if}\quad \theta^i \gt \theta_B \end{cases} $$

without proof. I am getting very tired of being told this without explanation. On page 391 of "Introduction to Electrodynamics" 3rd edition by "David J. Griffiths" is the same graph as that on the left of the image above and his reasoning is that

On the graph, a negative number indicates that the wave is $180^{\circ}$ out of phase with the incident beam.

Which again is useless to me and doesn't explain with any rigour the mathematical reasoning; which is the sort of explanation I am searching for.


If I had to guess I would say that the minus sign corresponding to a phase difference of $\pi$ would have to come from the dot product within the argument of the electric field:

$$\vec E_i=\vec E_{0_i}e^{i(\omega t - k_i\cdot \vec r)}\tag{1}$$ $$\vec E_r=\vec E_{0_r}e^{i(\omega t - k_r\cdot \vec r)}\tag{2}$$ $$\vec E_t=\vec E_{0_t}e^{i(\omega t - k_t\cdot \vec r)}\tag{3}$$

Dividing equation $(2)$ by $(1)$ I find that by definition $$r_p=\frac{E_r}{E_i}=\frac{\vec E_{0_r}}{\vec E_{0_i}}\frac{e^{i(\omega t - k_r\cdot \vec r)}}{e^{i(\omega t - k_i\cdot \vec r)}}=\frac{\vec E_{0_r}}{\vec E_{0_i}}e^{i \vec r(k_i-k_r)}$$

But I don't know how to proceed to show that if $$\frac{\vec E_{0_r}}{\vec E_{0_i}}e^{i \vec r(k_i-k_r)}\lt 0\tag{?}$$ then there is a phase difference of $\pi$ between $\vec E_i$ and $\vec E_r$.

Could anyone please help me complete the proof or show me if there is another way to prove that a negative reflection to incident ratio means a phase change of $\pi$?


I have already read similar questions on this site; like this, this and this popular question but they still don't answer my question here.


Update:

I have since been given an answer that mentions a special case of Euler's Identity; the answer states that

any minus sign can be reinterpreted as a phase shift of $\pi$, which you can bring into the field argument.

Using the inequality $(\mathrm{?})$ and Euler's formula means that $$\frac{\vec E_{0_r}}{\vec E_{0_i}}e^{i \vec r(k_i-k_r)}\lt 0\implies \vec r(k_i-k_r)=\pi\tag{4}$$

So how does $(4)$ show that any negative number has a phase difference of $\pi$ from the reflection to incident ratio?

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    $\begingroup$ This question shows research effort like very few I've seen; +1 $\endgroup$ – Alfred Centauri Aug 27 '17 at 0:56
  • $\begingroup$ Because if you multiply a vector by -1 it points in the opposite direction, i.e. is phase-shifted by $\pi$. $\endgroup$ – Rob Jeffries Aug 27 '17 at 20:30
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I don't suppose it would be satisfying for you if I recall everyone's favorite equation, $e^{i \pi}=-1$? So any minus sign can be reinterpreted as a phase shift of $\pi$, which you can bring into the field argument.

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  • $\begingroup$ Thank you for your answer and yes it is satisfying to recall arguably the most fascinating equation ever written. However that equation has a phase (shift) of $\pi$ for $-1$ only (at least I think). What about in the general case where we have a negative number $\ne -1$ (say $-3$ instead)? The inequality denoted by $(\mathrm{?})$ has to shown to have a phase difference of $\pi$ $\forall\, \Bbb Z_{\lt 0}$ . $\endgroup$ – BLAZE Aug 27 '17 at 1:57
  • $\begingroup$ Ah, I think I see your confusion. In general the reflection coefficient, r, is a complex number. If you write the complex r as $A*e^{i \theta}$, then A will give you the amplitude change, and &\theta& will give you the phase shift. This comes up when you consider real materials with loss, e.g. metals (which have refractive indexes that are almost all imaginary). So if $r=-0.5$, that's an amplitude change of a factor of 0.5, and a phase shift of $\pi$. $\endgroup$ – Gilbert Aug 27 '17 at 3:14
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    $\begingroup$ I should have also mentioned before, this is a good question. You have made an astute observation that there should be no physical reason for phase shifts upon reflection to be only zero or 180°. Indeed they are not, but for the textbook example of an idealized dielectric reflector. $\endgroup$ – Gilbert Aug 27 '17 at 3:29
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    $\begingroup$ Thank you, in your preceding comment, writing complex $\vec r$ as $Ae^{i \theta}=\dfrac{\vec E_{0_r}}{\vec E_{0_i}}e^{i \vec r(k_i-k_r)}$ so $A=\dfrac{\vec E_{0_r}}{\vec E_{0_i}}$ and $\theta=\vec r (k_i-k_r)$ but how does it follow from this that $\theta = \pi$ if $\dfrac{\vec E_{0_r}}{\vec E_{0_i}}e^{i \vec r(k_i-k_r)} \lt 0$? Can you see what it is that I'm trying to accomplish? $\endgroup$ – BLAZE Aug 27 '17 at 3:54
  • $\begingroup$ I turns out that for any real, negative number written as an amplitude multiplying a complex argument, the argument will be $\pi$. In the complex plane, numbers on the real axis have either an argument of 0 or $\pi$. Try it! $\endgroup$ – Gilbert Aug 27 '17 at 4:12
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You already got pretty close with your (?)-identity. First thing to realise is that electromagnetic wave scattering is elastic, so

$$k_i = k_r.$$

That simplifies (?) to $$\frac{\vec E_{0_r}}{\vec E_{0_i}} < 0.$$

So this is simply a condition on the complex amplitudes of the field. Let us decompose the amplitudes into magnitude and phase: $$\vec E_{0_i} = \vec A_{0_i} e^{i\phi_i},$$ $$\vec E_{0_r} = \vec A_{0_r} e^{i\phi_r}$$ where $\vec A_{0_i}$, $\vec A_{0_r}$ are real and positive. So condition (?) becomes

$$\frac{\vec A_{0_r}}{\vec A_{0_i}} e^{i(\phi_r-\phi_i)} <0$$

and hence

$$e^{i(\phi_r-\phi_i)} <0.$$

$\phi_r-\phi_i$ is the phase difference. Of course it could be anything, but again due to elasticity the reflection coefficient is always real. So the above factor only takes 2 values

$$e^{i(\phi_r-\phi_i)} \in \cases{+1 \\ -1}.$$

And this is what people mean by the phase shift being $\pi$ when the reflection coefficient is negative.

Two remarks:

  • I claimed that electromagnetic wave scattering is elastic. This is because we have made the implicit assumption that the refractive indices are real. If they are complex you have absorption in the system, the reflection coefficient becomes complex and $e^{i(\phi_r-\phi_i)} <0$ does not even make sense anymore since the left hand side is a complex number.
  • A picture always helps. Reflection is about waves coming in and going out. In 1D: enter image description here Then the meaning of phase shifts becomes clear.
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